Problem: Solve for $x$ : $ 5|x + 6| + 5 = -1|x + 6| + 7 $
Add $ {1|x + 6|} $ to both sides: $ \begin{eqnarray} 5|x + 6| + 5 &=& -1|x + 6| + 7 \\ \\ { + 1|x + 6|} && { + 1|x + 6|} \\ \\ 6|x + 6| + 5 &=& 7 \end{eqnarray} $ Subtract ${5}$ from both sides: $ \begin{eqnarray} 6|x + 6| + 5 &=& 7 \\ \\ { - 5} &=& { - 5} \\ \\ 6|x + 6| &=& 2 \end{eqnarray} $ Divide both sides by ${6}$ $ \dfrac{6|x + 6|} {{6}} = \dfrac{2} {{6}} $ Simplify: $ |x + 6| = \dfrac{1}{3}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 6 = -\dfrac{1}{3} $ or $ x + 6 = \dfrac{1}{3} $ Solve for the solution where $x + 6$ is negative: $ x + 6 = -\dfrac{1}{3} $ Subtract ${6}$ from both sides: $ \begin{eqnarray} x + 6 &=& -\dfrac{1}{3} \\ \\ {- 6} && {- 6} \\ \\ x &=& -\dfrac{1}{3} - 6 \end{eqnarray} $ Change the ${ - 6}$ to an equivalent fraction with a denominator of $3$ $ x = - \dfrac{1}{3} {- \dfrac{18}{3}} $ $ x = -\dfrac{19}{3} $ Then calculate the solution where $x + 6$ is positive: $ x + 6 = \dfrac{1}{3} $ Subtract ${6}$ from both sides: $ \begin{eqnarray} x + 6 &=& \dfrac{1}{3} \\ \\ {- 6} && {- 6} \\ \\ x &=& \dfrac{1}{3} - 6 \end{eqnarray} $ Change the ${ - 6}$ to an equivalent fraction with a denominator of $3$ $ x = \dfrac{1}{3} {- \dfrac{18}{3}} $ $ x = -\dfrac{17}{3} $ Thus, the correct answer is $x = -\dfrac{19}{3} $ or $x = -\dfrac{17}{3} $.